Answer
(a) See graph, and directions indicated.
(b) $\frac{(x-3)^2}{9}+\frac{y}{4}=1$ with $x\geq3$
Work Step by Step
(a) See graph, at t=0, the starting point is at (3,2), as t increases, the curve goes initially from left to right as shown in the figure and then turn around at $t=\pi/2$ and goes from right to left as shown in the bottom half of the curve.
(b) Rewrite the equations as $sin(t)=\frac{x-3}{3}, cos(t)=\frac{y}{2}$, use the Pythagorean Identity $sin^t+cos^t=1$ we have $(\frac{x-3}{3})^2+(\frac{y}{2})^2=1$ or $\frac{(x-3)^2}{9}+\frac{y}{4}=1$ with $x\geq3$ (which is the result of $0\leq t\leq\pi$)
